Figure rhom is a rhombus. and are the diagonals of the rhombus, as well as angle bisectors of the vertex angles, and they create four isosceles triangles: hom, mhr, rho, and omr. what is true about msr? it must be acute. it must be a right angle. it must be equal to mrh. it must be equal to rms.

∠MSR is right angle

Step-by-step explanation:

It is given that RHOM is a Rhombus

Also ΔHOM, ΔMHR, ΔRHO and ΔOMR are isosceles triangles

Let us take ΔMSR and ΔRSH

∠MRS =∠HRS            ( since it is given that the diagonal RO bisects ∠R)

∠RMS =∠RHS     ( since Δ MRH is isosceles triangle )

RS = RS        ( common side )

By AAS congruency rule ΔMSR ≅ ΔHSR

so we have

∠MSR=∠RSH  ( corresponding parts of congruent triangles are congruent)

also we have

∠MSR +∠RSH =180°   ( supplementary angles)

∠MSR +∠MSR=180°          ( since ∠MSR=∠RSH)

2∠MSR= 180°

∠MSR =90°

Hence ∠MSR is right angle

It must be a right angle.

Step-by-step explanation:

The figure attached shows the rhombus RHOM with RO and HM as diagonals and are the angle bisectors of the vertex angles.

Let S be the point where the diagonals RO and HM intersects each other.

ΔHOM, ΔMHR, ΔRHO, ΔOMR are four isosceles triangles in the given rhombus.

Since, Diagonals of a rhombus bisect each other at right angle.

Therefore, we have ∠MSR= 90°

That is, ∠MSR is a right angle.

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step-by-step explanation:

can you put the picture up?

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