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Anonymous05/14/26, 12:41No.16976028You teach, I learn :))
Can someone explain how to approach this geometry proof?:(( I tried... but I got stuck on part b
Problem:
Let ABCD be a rectangle. Let H be a point on diagonal AC such that BH ⟂ AC.
a) Prove that △ABH ∼ △ACB.
b) Let O be the intersection point of AC and BD. Let K be the midpoint of AB. Suppose BH intersects OK at G, and AG intersects OB at L.
Prove that LH ∥ AB.
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Anon05/14/26, 15:01No.16976070
>You teach, I learn :))
Okay, I'm your teacher. So take notes, yeah?
>Prove that △ABH ∼ △ACB.
angle BAH = θ = angle BAC (since A, C, and H are collinear)
angle AHB = π/2 = angle ABC (is given)
angle ABH = π/2 – θ = angle ACB (since α + β + γ = π)
Thus those two triangles have the same three angles.
Hence they're similar triangles.
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Anon05/14/26, 19:13No.16976153
The reason those lines have to be parallel is that the entire figure is obviously symmetrical about line KG. Your teacher probably wants a proof using triangle congruence, so what you need to do is work through the diagram finding triangles which you can show are congruent to their reflections across line KG, until you have enough to prove lines LH and AB parallel.