Question

In \( \square_{ ABCD } \), side \( BC \| \) side \( AD \). Diagonal \( AC \) and diagonal \( BD \) intersect in the point \( Q \). If \( AQ =\frac{1}{3} AC \), then show that \( DQ =\frac{1}{2} B Q \)

Answer 1

Answer:

Given: in quadrilateral ABCD,

AD ║ BC and AQ=1/3 AC

To prove: DQ=1/2 BQ

Proof:

Since, AQ=1/3 AC

⇒ AQ/AC = 1/3

Let, AQ = x and AC = 3x

Where x is any number,

⇒  CQ = AC - AQ = 3x - x = 2x

Now, In quadrilateral ABCD,

AD ║ BC

By the Alternative interior angle theorem,

∠QAD≅∠QCB ,

∠QDA≅∠QBC

By AA similarity postulate,

△ADQ∼△CBQ

Now, By the property of similar triangles,

DQ/BQ = AQ/CQ

DQ/BQ = x/2x

then; DQ = 1/2 BQ

​Hence, Proved 

 

Answer 2

∵ AQ = 1/3 AC

Let AQ = x

∵ AC = 3x

CQ = AC - AQ = 3x - x = 2x

Now,

 ∵ AD || BC

∠QAD = ∠ QCB

∠QDA = ∠QCB

∴ ∠ By AA similarity

Δ ADQ ~ Δ CBQ

\(\frac{DQ}{BQ}=\frac{AQ}{CQ}\)

⇒ \(\frac{BQ}{BQ}=\frac{x}{2x}\)

∴ DQ = 1/2 BQ