Question

prove in an isosceles triangle, angles opposite to equal sides are equal . In the given figure abc is a triangle AB = AC, BL is perpendicular to ac and CM is perpendicular to AE show that BL= CM. Also prove AM=AL​

Answer 1

$\fbox \green{Answer \: }$

Answer 2

Answer:

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A )see figure). Show that ΔABC ≌ ΔABD. What can you say about BC and BD?

Ans. In quadrilateral ABCD we have

AC = AD

and AB being the bisector of ∠A.

Now, in ΔABC and ΔABD,

AC = AD

[Given]

AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

∴ Using SAS criteria, we have

ΔABC ≌ ΔABD.

∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

∴ BC = BD.

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that

(i) ΔABD ≌ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Ans. (i) In quadrilateral ABCD, we have AD = BC and

∠DAB = ∠CBA.

In ΔABD and ΔBAC,

AD = BC

[Given]

AB = BA

[Common]

∠DAB = ∠CBA

[Given]

∴ Using SAS criteria, we have ΔABD ≌ ΔBAC

(ii) ∵ ΔABD ≌ ΔBAC

∴ Their corresponding parts are equal.

⇒ BD = AC

(ii) Since ΔABD ≌ ΔBAC

∴ Their corresponding parts are equal.

⇒ ∠ABD = ∠BAC.

3.