Advertisements
Advertisements
प्रश्न
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that : ED = EF

Advertisements
उत्तर
In ΔEFB and ΔEDB,
∠EFB = ∠EDB ( both are 900 )
EB = EB ( common side )
∠FBE = ∠DBE ( given )
ΔEFB ≅ ΔEDB (AAS congruence criterion)
⇒ EF = ED (cpct )
that is , Ed = EF.
APPEARS IN
संबंधित प्रश्न
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Explain, why ΔABC ≅ ΔFED.

In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In the given figure, prove that:
CD + DA + AB + BC > 2AC

In the given figure, prove that:
CD + DA + AB > BC

The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
In ∆ABC, AB = AC. Show that the altitude AD is median also.
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.
Prove that: ΔBEC ≅ ΔDCF.
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.
Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).
