Q5 of 31 Page 128

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that∠ B = ∠ C.

It is given in the question that:

AB = AC



In and


∠APB = ∠APC = 90o (AP is altitude)


AB = AC (Given)


AP = AP (Common)


Therefore,


By RHS axiom,



Thus,


∠B = ∠C (By c.p.c.t)


More from this chapter

All 31 →
3

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). Show that:

(i) Δ ABM ≅Δ PQN


(ii) Δ ABC ≅Δ PQR


 4

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

 1

Show that in a right angled triangle, the hypotenuse is the longest side.

 2

In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠ PBC <∠ QCB. Show that AC > AB.