Can someone explain how to approach this geometry proof?:(( I tried... but I got stuck on part bProblem: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.
>>16976028>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.
>>16976070I'll definitely take notes :Đ thank you for helping with part a
>>16976070I have a question :Đ my teacher said that if two triangles have two equal angles.. then they're automatically similar..is that true?
>>16976079Interior angles of a triangle always add up to 180 degrees. So if two angles are known, the third can be calculated.So yes, if two angles are the same, the third must also be the same as well.
>>16976080oh thank you :Đ!
>>16976028The 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.
>>16976153Drop two perpendicular lines from line AB that intersect points L and H respectively. If you can prove those two lines are of equal length, then you have proven LH is parallel to AB.
