Question

D is the midpoint of BC of ∆ ABC . From the point B, BE is perpendicular to the external bisector of BAC Prove that DE =( AB+AC)​

Answer 1

Answer:

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.