A geometry problem by Wildan Bagus Wicaksono

Geometry Level 1

Given the △ A B C \triangle ABC . The lines A D AD and B E BE are each perpendicular to the sides B C BC and A C AC and intersect at point H H . Find the value

( A H ⋅ H D B H ⋅ H E ) 10 { \left( \frac { AH\cdot HD }{ BH\cdot HE } \right) }^{ 10 }


The answer is 1.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

See the picture above. △ A H E \triangle AHE opposite to △ B H D \triangle BHD means ∠ A H E = ∠ B H D \angle AHE=\angle BHD . ∠ A E H = ∠ B D H = 90 \angle AEH=\angle BDH=90 and ∠ E A H = ∠ D B H \angle EAH=\angle DBH . This means that △ A H E \triangle AHE is similar to △ D B H \triangle DBH . Based on equivalent comparisons on two similar triangles:

A H B H = H E H D \frac { AH }{ BH } =\frac { HE }{ HD } or A H ⋅ H D = B H ⋅ H E AH\cdot HD=BH\cdot HE .

So, ( A H ⋅ H D B H ⋅ H E ) 10 = 1 10 = 1 { \left( \frac { AH\cdot HD }{ BH\cdot HE } \right) }^{ 10 }={ 1 }^{ 10 }=\boxed { 1 } .

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