Construction:
Step (1) Draw a line segment QR = 5 cm.
Step (2) At Q, draw QE such that \(\angle RQE\) = 40°.
Step (3) At Q, draw QF such that \(\angle EQF\) = 90^o
Step (4)Drawn a perpendicular bisector to QR, which intersects QF at 'O' and QR at G.
Step (5) With O as centre and OQ as radius, draw a circle
Step (6) From G marked arcs of radius 4.4 cm on the circle. Marked them as P and S.
Step (7) Joined QP and PR. Now \(\triangle\)PQR is the required triangle
Step (8) From P draw a line PN which is \(\bot \) to LR. LR meets PN at M.
Step (9) The length of the altitude is PM = 2.1cm

Construction:
Step 1: With center 'O' drawn a circle of radius 5 cm
Step 2: Drawn a line OP = 8 cm
Step 3: Draw a perpendicular bisector of OP which cuts OP at M
Step 4: With M as center and MO as radius drawn a circle which cuts previous circle at A and B.
Step 5: Joined AP and BP. AP and BP are the required tangents.
AP = BP = 6.2 cm.

Let in \(\Delta\)ABC , B Q and R are the foot of the perpendiculars drawn from the vertices A, B and C respectively.
\(\Delta B R C \sim \Delta B P A\)
\(\because \angle B R C=\angle B P A=90^{\circ}\)
\(\angle \mathrm{B} \text { is common }[\therefore \text { By AA criteria }]\)
\(\frac{B R}{B P}=\frac{B C}{B A}\)
\(\text { Similarly } \triangle A Q B \sim \triangle A R C\)
\(\frac{A Q}{A R}=\frac{A P}{A C}\)
\(\text { and } \Delta C P A \sim \Delta C Q B\)
\(\frac{C P}{C Q}=\frac{A C}{B C}\)
\(\text { Multiplying (1), (2) and (3), we have }\)
\(\frac{B R}{B P} \times \frac{A Q}{A R} \times \frac{C P}{C Q}=\frac{B C}{A B} \times \frac{A B}{A C} \times \frac{A C}{B C}=1\)
\(\therefore \text { By Ceva's theorem altitudes are concurrent. }\)

Let ABCD be the given rectangle. Let 'O' be the point within it. Join OA, OB, OC and OD.
Through O draw EOF || AB. Then ABFE is a rectangle.
In right triangles \(\triangle O E A \text { and } \triangle O F C \text {, }\) we have
\(\mathrm{OA}^{2} =\mathrm{OE}^{2}+\mathrm{AE}^{2} \text { and } \mathrm{OC}^{2}=\mathrm{OF}^{2}+\mathrm{CF}^{2} \)
\(\mathrm{OA}^{2}+\mathrm{OC}^{2} =\left(\mathrm{OE}^{2}+\mathrm{AE}^{2}\right)+\left(\mathrm{OF}^{2}+\mathrm{CF}^{2}\right) \)
\(\mathrm{OA}^{2}+\mathrm{OC}^{2} =\mathrm{OE}^{2}+\mathrm{OF}^{2}+\mathrm{AE}^{2}+\mathrm{CF}^{2} \)
\(\text { Now in right triangles } \triangle O F B \text { and } \triangle D O E \text { we have }\)
\(\mathrm{OB}^{2}= \mathrm{OF}^{2}+\mathrm{FB}^{2} \text { and } \mathrm{OD}^{2}=\mathrm{OE}^{2}+\mathrm{DE}^{2} \)
\(\mathrm{OB}^{2}+\mathrm{OD}^{2}=\left(\mathrm{OF}^{2}+\mathrm{FB}^{2}\right)+\left(\mathrm{OE}^{2}+\mathrm{DE}^{2}\right) \)
\(\mathrm{OB}^{2}+\mathrm{OD}^{2}= \mathrm{OE}^{2}+\mathrm{OF}^{2}+\mathrm{DE}^{2}+\mathrm{BF}^{2} \)
\(\mathrm{OB}^{2}+\mathrm{OD}^{2}= \mathrm{OE}^{2}+\mathrm{OF}^{2}+\mathrm{CF}^{2}+\mathrm{AE}^{2} \)
\([\because \mathrm{DE}=\mathrm{CF} \text { and } \mathrm{AE}=\mathrm{BF}] \)
\(\text { From }(1) \text { and }(2) \text { we get } \mathrm{OA}^{2}+\mathrm{OC}^{2}=\mathrm{OB}^{2}+\mathrm{OD}^{2}\)

Let D be the midpoint of AB. join CD.
Since ABC is a right angled triangle, \(A C B=90^{\circ}\)
\(\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}\)
\(\mathrm{AB}^{2}=(\sqrt{3} B C)^{2}+\mathrm{BC}^{2}\)
\([\because \mathrm{AC}=\sqrt{3} \mathrm{BC} \text { given }]\)
\(\mathrm{AB}^{2} =3 \mathrm{BC}^{2}+\mathrm{BC}^{2} \)
\(\Rightarrow \mathrm{AB}^{2} =4 \mathrm{BC}^{2} \)
\(\mathrm{AB} =2 \mathrm{BC} \)
\(\text { But } \mathrm{BD} =\frac{1}{2} \mathrm{AB} \)
\(\mathrm{AB} =2 \mathrm{BD} \)
\(\text { (1) and (2) } \Rightarrow \mathrm{BD}=\mathrm{BC}\)
We know that the midpoint of the hypotenuse of a right triangle is equidistance from the vertices
\(\therefore \mathrm{CD}=\mathrm{AD}=\mathrm{BD} \)
\(\Rightarrow \mathrm{CD}=\mathrm{BC} \)
\(\text { Thus in } \triangle A B C \text { we have } \mathrm{BD}=\mathrm{CD}=\mathrm{BC}\)
\(\therefore \triangle B C D \text { is equilateral }\)
\(\Rightarrow \therefore \angle A B C=60^{\circ}\)

In right angled triangles ACQ and PCB, we have
\(\mathrm{AQ}^{2} =\mathrm{AC}^{2}+\mathrm{CQ}^{2} \text { and } \)
\(\mathrm{PB}^{2} =\mathrm{PC}^{2}+\mathrm{CB}^{2} \)
\(\Rightarrow \mathrm{AQ}^{2}+\mathrm{BP}^{2} =\left(\mathrm{AC}^{2}+\mathrm{CQ}^{2}\right)+\left(\mathrm{PC}^{2}+\mathrm{CB}^{2}\right) \)
\(\Rightarrow \mathrm{AQ}^{2}+\mathrm{BP}^{2} =\left(\mathrm{AC}^{2}+\mathrm{BC}^{2}\right)+\left(\mathrm{PC}^{2}+\mathrm{QC}^{2}\right) \)
[By Pythagoras theorem we have
\(\left.A C^{2}+B C^{2}=A B^{2} \text { and } P C^{2}+Q C^{2}=P Q^{2}\right] \)
\(\therefore \ A Q^{2}+B P^{2}=A B^{2}+P Q^{2} \)

Given: \(\Delta\)ABC is a triangle in which the bisectors of \(\angle\)B and \(\angle\)C meet the sides AC and AB at D and E respectively.
To prove: AB = AC
Construction: join DE
Proof : In \(\Delta\)ABC, BD is the bisector of \(\angle\)B
\(\therefore \frac{A B}{B C}=\frac{A D}{D C}\)
\(\text { In } \triangle A B C, C E \text { is the bisector of } \angle C\)
\(\therefore \frac{A C}{B C}=\frac{A E}{B E}\)
\(\text { Now DE } \| \mathrm{BC}\)
\(\Rightarrow \frac{A E}{B E}=\frac{A D}{D C}\) [By Thales theorem]
From (1), (2) and (3), we have
\(\frac{A B}{B C}=\frac{A C}{B C} \Rightarrow A B=A C\)
Hence \(\Delta\)ABC is isosceles.

Given: In \(\Delta\)ABC,, AD is the median and DE and DF are the bisectors of \(\angle\)ADB and \(\angle\)ADC respectively
meeting AB and AC in E and F respectively.
To prove: EF || BC
Proof
In \(\Delta\)ADB, DE is the bisector of \(\angle\)ADB
\(\therefore \frac{A D}{D B}=\frac{A E}{E B}\)
\(\text { In } \triangle A D C, \text { DF is the bisector of } \angle A D C\)
\( \frac{A D}{D C}=\frac{A F}{F C} \)
\(\Rightarrow \frac{A D}{D B}=\frac{A F}{F C} \)
\([\because \mathrm{AD} \text { is the median } \mathrm{BD}=\mathrm{DC}]\)
From (1) and (2), we get
\(\frac{A E}{E B}=\frac{A F}{F C}\)

\(\text { In } \triangle A O B, O D \text { is the bisector of } \angle A O B\)
\(\frac{O A}{O B}=\frac{A D}{D B}\)
\(\text { In } \triangle B O C, O E \text { in the bisector of } \angle B O C\)
\(\frac{O B}{O C}=\frac{B E}{E C}\)
\(\text { In } \triangle C O A, O F \text { is the bisector of } \angle C O A\)
\(\frac{O C}{O A}=\frac{C F}{F A}\)
Multiplying the corresponding sides of (1), (2) and (3) we get
\(\frac{O A}{O B} \times \frac{O B}{O C} \times \frac{O C}{O A}=\frac{A D}{D B} \times \frac{B E}{E C} \times \frac{C F}{F A}\)
\(1=\frac{A D}{D B} \times \frac{B E}{E C} \times \frac{C F}{F A}\)
\(\mathrm{DB} \times \mathrm{EC} \times \mathrm{FA}=\mathrm{AD} \times \mathrm{BE} \times \mathrm{CF} \)
\(\mathrm{AD} \times \mathrm{BE} \times \mathrm{CF}=\mathrm{DB} \times \mathrm{EC} \times \mathrm{FA} \)

Steps of construction:
l. Drawn BC = 8 cm.
2. Drawn the perpendicular bisector of BC which intersects BC at D.
3. Marked a point A on the above perpendicular such that DA = 4 cm.
4. joined AB and AC. Thus \(\Delta\)ABC is the required isosceles triangle.
5. Then drawn array BX such that \(\angle\)CBX is an acute angle.
5. On BX, marked three points (since \(\left.1 \frac{1}{2}=\frac{3}{2}\right) X_{1}X_{2} \text { and } X_{3} \text { such that } B X_{1}=X_{1} X_{2}=X_{2} X_{3}\)
7. Joined X₂ to C.
8. Drawn a line through X₃, parallel to X₂C and intersecting BC extended to C'.
9. Drawn a line through C' parallel to CA intersecting BA extended at A'. Thus \(\Delta\)A'BC' is the required triangle.

We have a square ABCD, whose diagonal AC.
Equilateral triangle \(\Delta\)BQC is described on the side BC and another equilateral \(\Delta\)APC is described on the diagonal AC.
\(\therefore\) All equilateral triangles are similar.
\(\triangle A P C \sim \triangle B Q C\)
\(\therefore\)The ratio of their areas is equal to the square of the ratio of their corresponding sides.
\(\frac{\text { area }(\Delta \mathrm{APC})}{\text { area }(\triangle \mathrm{BOC})}=\left(\frac{A C}{B C}\right)^{2}\)
Since the length of a diagonal of a square \(=\sqrt{2} \times \operatorname{side}\)
\(\mathrm{AC}=\sqrt{2} \times B C\)
From (1) and (2) we have
\(\frac{\text { area }(\Delta \mathrm{APC})}{\operatorname{area}(\Delta \mathrm{BQC})}=\left(\frac{\sqrt{2} \mathrm{BC}}{B C}\right)^{2}\)
\(=(\sqrt{2})^{2}=2\)
\(\therefore \text { area }(\Delta B Q C)=\frac{1}{2} \operatorname{area}(\triangle A P C)\)

We have \(\triangle A B E \cong \triangle \text { ACD, }\)
Their corresponding parts are equal.
\(\mathrm{AB} =\mathrm{AC} \)
\(\mathrm{AE} =\mathrm{AD} \)
\(\therefore \frac{A B}{A C} =\frac{A E}{A D} \Rightarrow \frac{A B}{A E}=\frac{A C}{A D} \)
\(\Rightarrow \frac{A B}{A D} =\frac{A C}{A E} \quad[\because \mathrm{AE}=\mathrm{AD}] \)
\(\text { Now in } \triangle A D E \text { and } \triangle A B C\)
\(\frac{A B}{A D} =\frac{A C}{A E} \)
\(\angle D A E =\angle B A C \)
\(\text { Using SAS similarity criteria we have }\)
\(\triangle A D E \sim \triangle A B C\)