Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?
In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
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 two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
In the given figure: AB//FD, AC//GE and BD = CE;
prove that:
- BG = DF
- CF = EG
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 the following diagram, AP and BQ are equal and parallel to each other. 
Prove that: AB and PQ bisect each other.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
