Problem 5

In tetrahedron

ABCD

\angle BDC=90^\circ

. The foot

H

of the perpendicular from

D

to plane

ABC

is the intersection of the altitudes of

\triangle ABC

. Prove that

(AB+BC+CA)^2\le6(AD^2+BD^2+CD^2)

. For what tetrahedra does equality hold?

Step 2 of 4: Force the remaining right angles

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BC^2=BD^2+CD^2\implies CE^2=DE^2+CD^2\implies \angle CDE=90^\circ

Detailed analysis

The given angle $BDC$ is right, so $BC^2=BD^2+CD^2$ . Comparing this with the preceding identity gives $CE^2=DE^2+CD^2$ , hence angle $CDE$ is right. Since $CDB$ is also right, $CD$ is perpendicular to the two intersecting lines $DB$ and $DE$ in plane DAB; thus $CD\perp$ plane DAB and angle $CDA$ is right. By the same argument, angle $ADB$ is right.