Plane parallel to the base of a triangular pyramid

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A triangular pyramid $ABCD$ is given. A plane parallel to the plane $(ABC)$ intersects $DA,DB$ and $DC$ at $A_1,B_1$ and $C_1,$ respectively. Show that $\dfrac{DA_1}{DA}=\dfrac{DB_1}{DB}=\dfrac{DC_1}{DC}$.I think I am supposed to use Thales theorem for the given equality but I am not sure how to begin the solution. Thank you in advance!

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1 Answer

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Consider a side of the pyramid, say $\triangle ABD$. $\frac{DA_1}{DA}=\frac{DB_1}{DB}$ then follows from intercept theorems.

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