How prove that $|QA| < |QC|$ in triangle?

Question

$ABC$ is a triangle with a right angle at $A$, and $|AB|$ > $|AC|$. The point $D$ is defined so that $BCD$ is equlateral and $AD$ intersects $BC$ at $P$. The point $Q$ is defined so that $QDP$ is equilateral and $QP$ intersects with $DB$. How prove that $|QA| < |QC|$?

Answer 1

One approach, which I admit seems like a bit of work is to treat the points as vectors and try to compute each location.

You can:

• Take $A$ as the origin.
• Take $B$ and $C$ as defining an orthogonal basis (not orthonormal)
• Define $D$ as $\mu$$B$+$\lambda$$C$

Then $D$ must solve the equations $(D-C)^2=(B-C)^2=(D-B)^2$.

These should be solvable to give $\mu$ and $\lambda$ in terms of the ratio $|B|/|C|$.

One ought then to proceed to determine $P$, which is of the form:

$P = B + \gamma(B-C)$

and then in turn determine $Q$ similarly, being:

$Q = \phi B+\rho C$

Having done all that, the requirement is to show that:

$Q^2< (Q-C)^2$

or equivalently that

$\rho < 1/2$