Question

ABCD is a square. Diagonals AC and BD intersect each other at O. Bisector of∠BAC meet BO at P and BC at Q. Prove that OP = ½ CQ

Answer 1

Concept we will be using:
i) Each of the angles in a square is 90 degrees.
ii) Diagonals of a square bisect each other at 90 degrees.
iii) Diagonals of a square bisect each pair of opposite angles.
iv) Sides opposite to congruent angles are congruent.
v) Triangle angle bisector theorem ( Please find the attachment)
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Solution:

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.

R.T.P: $OP = \frac{1}{2}CQ$

Proof:
The diagonal BD bisects∠ABC and ∠ABC =90 because angle of a square.
$\angle{ABD}= \frac{1}{2}\angle{ ABC}= \frac{1}{2} \times90=45^o$


The diagonal AC bisects∠BAD and ∠BAD =90 because angle of a square.
$\angle{BAC}= \frac{1}{2}\angle{ BAD}= \frac{1}{2} \times90=45^o$

AQ bisects ∠BAC
$\angle{BAQ}= \frac{1}{2}\angle{ BAC}= \frac{1}{2} \times45=22.5^o$



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:
$\frac{OP}{BP} = \frac{AO}{AB}\text{ .......(ii)}$


In ΔACB, AQ is the angle bisector.
By, triangle angle bisector theorem, we have:
[tex]\frac{CQ}{BQ} = \frac{AC}{AB} \\ \text{Since, diagonals AC and BD bisect each other at O, then AC=2AO} \\ \text{Plugging in AC = 2AO}:\\ \frac{CQ}{BQ} = \frac{2AO}{AB} \\\text{Plugging in } \frac{AO}{AB} = \frac{OP}{BP} \text{ from (ii) into (iii):}\\ \frac{CQ}{BQ} = 2\times\frac{AO}{AB}\\ \frac{CQ}{BQ} = 2\times \frac{OP}{BP}\\ \text{Plugging in BQ=BP from (i):}\\ \frac{CQ}{BP} = 2\times \frac{OP}{BP}\\ \text{Cancel out BP from both sides:}\\ CQ= 2OP \\ \text{Divide both sides by 2:}\\ \frac{CQ}{2}= \frac{2OP}{2}\\ \frac{1}{2}CQ=OP\\ OP=\frac{1}{2}CQ \text{ (Proved)}[/tex]

Answer 2

Proof of the question is given with the attachment.
But we need to have some pre-requisite knowledge of properties of diagonals of square and the ∠'s : 
1. Diagonals of square bisect each other.
2. Diagonals of square are the ∠bisectors.

so from the above properties we have
∠BAC=∠CAD=∠ADB=∠BDC=∠ACD=∠ACB=∠ABD=∠CBD=45°
∠BAQ=∠CAQ=22.5°
now refer to the picture 
and please mark the answer as brainliest