Question

In trapezoid ABCD with bases AB and DC , diagonals intersect at point O. Prove that: AO/BO = AC/BD = CO/DO

Answer 1

Step-by-step explanation:

Given: ABCD is a trapezoid with bases AB and DC, and the diagonals intersect at point O.

To prove: $\frac{AO}{BO}=\frac{AC}{BD}=\frac{CO}{DO}$

Construction: Draw EF parallel to both AB and DC which passes through o.

Proof: In ΔADC, we have

EO is parallel to DC( because EF is parallel to DC)

Therefore, $\frac{AE}{DE}=\frac{AO}{CO}$ (Line drawn parallel to one side of triangle, intersects the other two sides at different points, thus dividing the other two in equal ratio.)                                    (1)

Similarly, in ΔDBA,

EO is parallel to AB

Thus, $\frac{AE}{DE}=\frac{BO}{DO}$                            (2)

From equations (1) and (2),

$\frac{AO}{BO}=\frac{CO}{DO}$

Also, diagonals bisect each other at O, therefore

$\frac{AO}{BO}=\frac{AC}{BD}=\frac{CO}{DO}$

Hence proved.