In $\Delta ABC$, $\angle BAC=90{}^\circ $, $AD$ is perpendicular to $BC$ and $M$ is the mi

Question

In $\Delta ABC$, $\angle BAC=90{}^\circ $, $AD$ is perpendicular to $BC$ and $M$ is the midpoint of $AD$. $BMG$ and $EGF$ are straight lines where $EGF$ is perpendicular to $BC$. Show that $\Delta BGF$ is similar to $\Delta BMD$, $\Delta BEG$ is similar to $\Delta BAM$. Hence, show that $EG=GF$. Prove that $GF^2=AG\times GC$.

Accepted Answer

(i)(a) Similar; (i)(b) Similar; (ii) EG = GF; (iii) GF^2 = AG x GC (proved).

Proofs in Plane Geometry - Similarity

Proofs in Plane Geometry - Similarity

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