Q2 of 31 Page 128

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC


(ii) AD bisects ∠ A

It is given in the question that:

AD is an altitude and AB = AC



(i) In


∠ADB = ∠ADC = 90o


AB = AC (Given)


AD = AD (Common)


Therefore,


By RHS axiom,




(By c.p.c.t)




(ii) ∠BAD = ∠CAD (By c.p.c.t)


Thus,


Ad bisects ∠A


More from this chapter

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8

Show that the angles of an equilateral triangle are 60° each.

 1

Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that

(i) Δ ABD ≅Δ ACD


(ii) Δ ABP ≅Δ ACP


(iii) AP bisects ∠ A as well as ∠ D.


(iv) AP is the perpendicular bisector of BC.


 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.