Q2 of 31 Page 123

In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ ABC is an isosceles triangle in which AB = AC.

It is given in the question that:

AD is the perpendicular bisector of BC


To show: AB = AC


Proof: In


AD = AD (Common)


∠ADB = ∠ADC


BD = CD (AD is the perpendicular bisector)


Therefore,


By SAS congruence axiom,



AB = AC (By c.p.c.t)


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8

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B(see Fig. 7.23). Show that:

(i) Δ AMC ≅Δ BMD


(ii) ∠DBC is a right angle.


(iii) Δ DBC ≅Δ ACB


(iv) CM =AB

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In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:

(i) OB = OC


(ii) AO bisects ∠ A

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ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

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ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32) Show that:

(i) Δ ABE ≅Δ ACF


(ii) AB = AC, i.e., ABC is an isosceles triangle