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Feet of perpendiculars to diagonal in cyclic quadrilateral
jl_   2
N Apr 23, 2025 by lw202277
Source: Malaysia IMONST 2 2023 (Primary) P6
Suppose $ABCD$ is a cyclic quadrilateral with $\angle ABC = \angle ADC = 90^{\circ}$. Let $E$ and $F$ be the feet of perpendiculars from $A$ and $C$ to $BD$ respectively. Prove that $BE = DF$.
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jl_
Apr 23, 2025
lw202277
Apr 23, 2025
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Feet of perpendiculars to diagonal in cyclic quadrilateral
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Source: Malaysia IMONST 2 2023 (Primary) P6
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jl_
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jl_
#1 h  Apr 23, 2025, 10:42 AM
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Suppose $ABCD$ is a cyclic quadrilateral with $\angle ABC = \angle ADC = 90^{\circ}$. Let $E$ and $F$ be the feet of perpendiculars from $A$ and $C$ to $BD$ respectively. Prove that $BE = DF$.
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navier3072
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navier3072
#2 h  Apr 23, 2025, 11:49 AM
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Let $\angle BCA= \theta$ and $\angle DAC=\alpha$. Thus, $\angle BDA= \theta$ and $\angle DBC=\alpha$. Thus, \[BF=BC \cos(\alpha)=AC \cos(\theta) \cos(\alpha)\]\[DE=AD \cos(\theta)=AC \cos(\alpha) \cos(\theta)\]
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lw202277
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lw202277
#3 h  Apr 23, 2025, 1:29 PM
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Let $M$ be the midpoint of $BD$ and $H$ be the orthocenter of $ABC$. $BE = DF$ is equivalent to $M$ being the midpoint of $EF$.

By reflecting the orthocenter, $M$ is the midpoint of $CH$. We have $HM=CM$ and $\angle HME = \angle CMF$ so $\triangle MHE \cong \triangle MCF$ by HL. This implies that $ME = MF$.
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