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We can prove congruency this way. Let the intersection point be X so we have 4 triangles: AXC, AXD, CXB, BXD. Together AXC and AXD make triangle ADC, and CXB and BXD make BDC.

AXC and CXB are congruent because they share side CX, angle CXA=CXB=90°, AX=XB because CD is the perpendicular bisector of AB. Also, by Pythagoras, the hypotenuses AC and CB are equal. We can apply the same reasoning to triangles AXD and BXD, where the common side is XD, angles AXD and DXB=90°, and AX=XB. So we have two pairs of congruent triangles, so the combined triangles ADC and BDC are congruent.

We also have AC=CB and AD=DB (hypotenuses of the smaller triangles) and CD is common, so we have 3 sides SSS of the two triangles ADC and BDC proved congruent. 

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