A geometry problem by A Former Brilliant Member

Geometry Level 2

In the above triangle, A D AD is an altitude. Point E E lies on the altitude. Find ( A C ) 2 + ( E B ) 2 − ( C E ) 2 − ( A B ) 2 . (AC)^2+(EB)^2-(CE)^2-(AB)^2.

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A Former Brilliant Member by A Former Brilliant Member

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1 solution

Apply pythagorean theorem on △ A D C : \triangle ADC: ( A C ) 2 = ( C D ) 2 + ( A D ) 2 (AC)^2=(CD)^2+(AD)^2 ⟹ \implies ( A D ) 2 = ( A C ) 2 − ( C D ) 2 (AD)^2=(AC)^2-(CD)^2

Apply pythagorean theorem on △ E D C : \triangle EDC: ( C E ) 2 = ( C D ) 2 + ( D E ) 2 (CE)^2=(CD)^2+(DE)^2 ⟹ \implies ( D E ) 2 = ( C E ) 2 − ( C E ) 2 (DE)^2=(CE)^2-(CE)^2

By subtraction: ( A D ) 2 − ( D E ) 2 = ( A C ) 2 − ( C E ) 2 (AD)^2-(DE)^2 = (AC)^2 - (CE)^2 ( 1 ) \color{#D61F06}(1)

Apply pythagorean theorem on △ A D B : \triangle ADB: ( A B ) 2 = ( A D ) 2 + ( D B ) 2 (AB)^2 = (AD)^2+(DB)^2 ⟹ \implies ( A D ) 2 = ( A B ) 2 − ( D B ) 2 (AD)^2=(AB)^2-(DB)^2

Apply pythagorean theorem on △ E D B : \triangle EDB: ( E B ) 2 = ( D E ) 2 + ( D B ) 2 (EB)^2=(DE)^2+(DB)^2 ⟹ \implies ( D E ) 2 = ( E B ) 2 − ( D B ) 2 (DE)^2 =(EB)^2-(DB)^2

By subtraction: ( A D ) 2 − ( D E ) 2 = ( A B ) 2 − ( E B ) 2 (AD)^2-(DE)^2=(AB)^2-(EB)^2 ( 2 ) \color{#D61F06}(2)

Equate ( 1 ) \color{#D61F06}(1) and ( 2 ) \color{#D61F06}(2) .

( A C ) 2 − ( C E ) 2 = ( A B ) 2 − ( E B ) 2 (AC)^2 - (CE)^2=(AB)^2-(EB)^2

Since the left side of the equation is equal to the right side of the equation,

( A C ) 2 + ( E B ) 2 − ( C E ) 2 − ( A B ) 2 = (AC)^2+(EB)^2-(CE)^2-(AB)^2= 0 \boxed{0}

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