Question

In the given figure, ABCD and AEFG are 2 squares. Prove that ∆ACF ~ ∆ADG.

Answer 1

read those 2 pic attatch,ents carefully, u ll understand

Answer 2

Given:

ABCD and AEFG are squares

By definition of square

AB=BC=CD=AD,AE=EF=FG=AG

Angle A=Angle B=Angle C=Angle D=90 degrees

Angle AGF=Angle GFE=Angle AEF=Angle GAE=90 degrees

To prove that

(i)$\frac{AF}{AG}=\frac{AC}{AD}$

(ii)$\triangle ACF$$\sim \triangle ADG$

Proof:

Construction: Draw AC

(i)AC and AF are diagonals of square ABCD and AEFG.

We know that

Diagonal of square=$\sqrt{2}\times$side of square

Using the formula

$AF=\sqrt{2}AG$

$AC=\sqrt{2}AD$

$\frac{AF}{AC}=\frac{\sqrt{2}AG}{\sqrt{2}AD}$

$\frac{AF}{AC}=\frac{AG}{AD}$

$\frac{AF}{AG}=\frac{AC}{AD}$

Hence, proved.

(ii)Angle AGF=90 degree

AG and GF are equal sides

Therefore, Angle GAF=Angle GFA=45 degrees

Because sum of angles of triangle =180 degrees

Diagonal AC divides the angle in to two equal parts

Therefore, Angle CAD=Angle CAB=45 degrees

Let

Angle FAC=x

Angle GAC=45-x

Angle GAD=Angle CAD-Angle CAG=45-(45-x)=45-45+x=x

Angle FAC=Angle GAD=x

$\frac{AF}{AG}=\frac{AC}{AD}$

$\triangle ACF\sim \triangle ADG$

By SAS similarity postulate

Hence, proved.