Question

Triangle ABCD side BC parallel side AD SIde AC and side BD intersect in point Q if AQ=1/3AC then show that DQ=1/2BQ​

Answer 1

Given: In quadrilateral ABCD, side BC parallel side AD SIde AC and side BD intersect in point Q.

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

[CQ = 2x]

Now, In quadrilateral ABCD,

AD ║ BC

By the Alternative interior angle theorem,

Angle QAD ≈ Angle QCB

Angle QDA ≈ Angle QBC

[By AA similarity postulate]

∆ADQ ~ ∆CBQ

Now, By the property of similar triangles,

DQ/BQ = AQ/CQ

DQ/BQ = x/2x

DQ = 1/2BQ

[Hence, it is proved.]