Q14 of 97 Page 205

Prove that the angles opposite to equal sides of a triangle are equal

We have a triangle PQR where PS is the bisector of ∠ P

Now in ∆PQS and ∆PSR, we have:

PQ = PR (Given)

PS = PS (Common)

∠ QPS = ∠ PRS (As PS is the bisector of ∠ P)

∴ By SAS congruence rule

∆PQS ≅ ∆PSR

∠ Q = ∠ R (By Congruent parts of congruent triangles)

Hence, it is proved that the angles opposite to equal sides of a triangle are equal

In ∆ABC, AB=AC. Side BA is produced to D such that AD=AB.

Prove that ∠BCD=90°.

In the given figure, it is given that AD=BC and AC=BD.

Prove that ∠CAD=∠CBD and ∠ADC=∠BCD.

In an isosceles ∆ABC, AB=AC and the bisectors of ∠B and ∠C intersect each other at O. Also, O and A are joined.

Prove that: (i) OB=OC (ii) ∠OAB=∠OAC

Prove that, of all line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest