Question

ABCD is a square. The bisector of DBC cuts AC and CD at E and F respectively. Prove that BF × CE =

BE × DF.​

Answer 1

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Step-by-step explanation:Here is your answer,

Given: ABCD is a square, AC and BD are the two diagonals which intersect each other at O. Also, given that  the angle bisector of ∠BAC meet BO at P and BC at Q.c

R.T.P: OP = \frac{1}{2}CQ  

Proof:

The diagonal BD bisects∠ABC and ∠ABC =90 because angle of a square.

\angle{ABD}= \frac{1}{2}\angle{ ABC}= \frac{1}{2} imes90=45^o

The diagonal AC bisects∠BAD and ∠BAD =90 because angle of a square.

\angle{BAC}= \frac{1}{2}\angle{ BAD}= \frac{1}{2} imes90=45^o

AQ bisects ∠BAC

\angle{BAQ}= \frac{1}{2}\angle{ BAC}= \frac{1}{2} imes45=22.5^o

In ΔABP,  

∠BAP = ∠BAQ =22.5 degrees ( Since A,P and Q lies on the same straight line)

∠ABP=∠ABD=45 degrees (Since B, P and D lies on the same straight line)

Sum of angles in a triangle is 180 degrees.

Therefore,  

∠BAP +∠ABP +∠APB=  180

⇒ 22.5+45+∠APB=  180

⇒ 67.5+∠APB=  180

⇒ ∠APB=  180-67.5

⇒ ∠APB=  112.5

In ΔBPQ,

∠BPQ= 180-∠APB =180-112.5=67.5 ( Since ∠BPQ and ∠APB constitute a straight line, then ∠BPQ and ∠APB must be supplementary)  

∠PBQ= ∠DBC=∠ABD =45 degrees (Since, BD bisects ∠ABC )

Sum of angles in a triangle is 180 degrees.

Therefore,  

∠BPQ +∠PQB +∠BQP=  180

⇒ 67.5 + 45 +∠BQP=  180

⇒ 112.5+∠BQP=  180

⇒ ∠BQP=  180-112.5

⇒ ∠BQP=  67.5

Therefore,  

∠BPQ = ∠BQP = 67.5 degrees

BQ is the side opposite to ∠BPQ

And, BP is the side opposite to ∠BQP  

We know that sides opposite to equal angles are equal.  

Then, BP = BQ......(i)  

In ΔAOB, AP is the angle bisector.

By, triangle angle bisector theorem, we have:

Answer 2

Step-by-step explanation:

this will help u

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