I do this later if no one does it yet, we need logical proofs for this

Part ii should be simply quoting the similarity of triangles length ratios and then cross-multiplying the result.

BAC similar to ADC
BAC similar to BDA

So ADC similar to BDA

Am I right?

I can’t do Part 2

Normally I would assign angles such as letting one angle be theta. I then proceed to work out that there will be a bunch of theta and 90 - theta.

For proving similar triangles, we have common right angle and another common angle. So all angles r same. I can’t do part 2

Let ABD = theta

Then,
BAD = 90 - theta (considering triangle ABD)
ACD = 90 - theta (considering triangle ABC)
CAD = theta (considering triangle ACD)

We can then say that
- ABD = CAD = theta
- BAD = ACD = 90 - theta
- ADB = CDA = 90 (they are side by side 90 degrees angle)

We can quote any of these two for the similarity of ABD and CAD by the two angles proof (AA test of similarity)

Wait, I need to see the diagram again, probably need another pair of similar triangles (all three are similar)

I try with theta

Hang on, I forgot to mention that I am in a class, I will complete this for you by tonight

Ok.Sure. Thanks

Let me see...

Since triangles ABD and CAD are similar,
AB/CA = AD/CD = BD/AD (statement 1)

Since triangles ABD and CBA are similar,
AB/CB = AD/CA = BD/BA (statement 2)

Since triangles CAD and CBA are similar,
CA/CB = CD/CA = AD/BA (statement 3)

We need to prove that BD/CD = AB²/AC².

From statement 2,
AB/CB = BD/BA
AB² = BC x BD

From statement 3,
CA/CB = CD/CA
AC² = BC x CD

Dividing statement 2 by statement 3,

AB²/AC² = (BC x BD) / (BC x CD)
AB²/AC² = BD/CD

Great! Thanks a lot Mr. Eric.