MHB Prove that triangle BAD is isosceles and....

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The discussion focuses on proving that triangle BAD is isosceles and establishing the relationship AC² + AD² = 4AB². Participants emphasize the use of the midpoint theorem, noting that point B is the midpoint of CD, which leads to the conclusion that AB equals BD. The congruence of triangles CBE and ABE is also highlighted, as they share equal angles and sides, reinforcing that CB equals BA. The conversation suggests that understanding the midpoint theorem is crucial for solving the problem effectively. Overall, the collaborative effort aims to clarify the geometric relationships within triangle ADC.
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Problem

In the $\triangle ADC$ , $\angle DAC$ or angle $A$ is a right angle, E is the midpoint of AC . The perpendicular drawn to $AC$ from $E$ meets $DC$ at $B$

i.Drawn the given information in a figure & prove that $\triangle BAD$ is isosceles

ii. $AC^2+AD^2=4AB^2$

Diagram


View attachment 6047

Where do I need help

In proving that $\triangle BAD$ is isosceles & $AC^2+AD^2=4AB^2$

This is the first time I attempt to do this kind of a problem (Clapping)

Many Thanks :)
 

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i) Start from the fact that B is the midpoint of CD (midpoint theorem).

ii) Start with the fact that 2AB = CD.
 
greg1313 said:
i) Start from the fact that B is the midpoint of CD (midpoint theorem).

ii) Start with the fact that 2AB = CD.

Thank you very much (Happy)

ii) Start with the fact that 2AB = CD

May i know the theorem or the case please :)

So $CB=BD$ (Converse of midpoint theorem)

& as $2AB = CD$ | $\therefore AB= \frac{1}{2} CD$ | $\frac{1}{2} CD $= CB or BD

$BA= BD$ | $\therefore \triangle BAD $ is isosceles

Correct?

In the first instance is it the Converse of the Midpoint theorem

The straight line through the midpoint of one side of a triangle and parallel to another side, bisects the third side.

or

the Midpoint theorem

The straight line segment through the midpoints of two sides of a triangle is
parallel to the third side and equal in length to half of it
 
Let's start with i) because we may use the results of i) to prove ii).

I'd like you to consider the problem carefully and decide which aspect of the midpoint theorem applies to which part of i). Post back with your answer and include your reasoning.

One way to solve i): consider a rectangle with diagonal BD and a rectangle with diagonal AB. Can you, using the midpoint theorem, show that these two rectangles are congruent? What famous theorem tells us that AB = BD?
 
greg1313 said:
Let's start with i) because we may use the results of i) to prove ii).

I'd like you to consider the problem carefully and decide which aspect of the midpoint theorem applies to which part of i). Post back with your answer and include your reasoning.

One way to solve i): consider a rectangle with diagonal BD and a rectangle with diagonal AB. Can you, using the midpoint theorem, show that these two rectangles are congruent? What famous theorem tells us that AB = BD?

Hey greg1313 :)

The triangle $CBE$ is equal to $ABE$ because they have equal angle in $E$ and equal sides $CE=EA$ and $BE=BE$.

So now what I have is $CB=BA$

Then the theorem should be the converse of the midpoint theorem

The straight line through the midpoint of one side of a triangle and parallel to another side, bisects the third side.

Many Thanks :)
 
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