# Thread: Prove equal area of a triangle and a parallelogram & the angle is equal a 90 degree

1. Workings

$\triangle ADE \cong \triangle CFE \left(AAS\right)$

$\angle AED = \angle CEF$( vertically opposite angles )
$\angle CFE= \angle EDA$( alternate angles )
$AE=EC$( E midpoint )

$ii.$ADCF is a parallelogram because diagonals bisect each other.

Where is help needed

How should the fact that area of $\triangle ABC$ is equal to parallelogram ADCF be proved ?

Show that if $DE=AE$, then $\angle ADC=90^{\circ}$

2. What is the area of triangle ADC in relation to the area of triangle ABC?

Originally Posted by greg1313
What is the area of triangle ADC in relation to the area of triangle ABC?
The triangle ADC exactly one half of the area of the triangle ABC

and the parallelogram ADCF is twice the area of triangle ADC and triangle ABC is twice the area of triangle ABC

Now what remains is,

Show that if $DE=AE$, then $\angle ADC=90^{\circ}$

Many THanks

4. What do the measures of two opposite angles in a cyclic quadrilateral sum to?

Originally Posted by greg1313
What do the measures of two opposite angles in a cyclic quadrilateral sum to?
180 degrees , But a cyclic quadrilateral is not to be seen here

6. Look again.

Originally Posted by greg1313
Look again.
A little bit confused here,

A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle.

But here there is no circle around the figure

Many Thanks

8. Originally Posted by mathlearn
Now what remains is,

Show that if $DE=AE$, then $\angle ADC=90^{\circ}$
As an alternative to greg1313's suggestion, what can you say about a parallelogram whose diagonals are equal?

9. Originally Posted by mathlearn
A little bit confused here,

A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle.

But here there is no circle around the figure

Many Thanks
The circle doesn't have to be there. One way of describing a cyclic quadrilateral is "A cyclic quadrilateral is a quadrilateral around which one can draw a circle that touches all four vertices". Now can you make progress with my hint (if you wish to do so)?

Originally Posted by greg1313
The circle doesn't have to be there. One way of describing a cyclic quadrilateral is "A cyclic quadrilateral is a quadrilateral around which one can draw a circle that touches all four vertices". Now can you make progress with my hint (if you wish to do so)?
That's what was exactly missing here,

AFD+DFC=AFC

dividing both sides by 2

Originally Posted by Opalg
As an alternative to greg1313's suggestion, what can you say about a parallelogram whose diagonals are equal?
Taking it the alternative way,

Both squares and rectangles have their diagonal equal & All angles of them are equal (90°)

It is a rectangle in this case as in a rectangle opposite sides are equal as in a parallelogram